# Uncanny triangles

Thomas, Jane and Anna were drawing right angled triangles on squared paper. Their triangles had two sides which were an exact number of squares long and could not be longer than $15$ squares. These are Jane's triangles:

They were calculating the areas of the triangles.

"I've got one triangle where the area and the sum of the lengths of the two shorter sides come to exactly the same number!" exclaimed Anna, "Look, it's that one!"

Thomas looked at his work. "How uncanny - but so have I! But look at it. It's quite a different shape from yours."

What were the measurements of the triangles they had drawn?

Where is a good place to start?

Have you looked at some of the examples on the sheet?

How many different triangles do you think you can draw with the lengths of the two perpendicular and horizontal sides making any total up to $15$? Can you list some? How do you know you can list them all without missing any out?

Have you thought of trying the "family" of triangles with a height of one?

Do you need to draw all the triangles in a family to see if any meet the requirements? How can you tell?

Julian from Hastingsbury Upper School approached this problem using algebra:

Assuming that all the triangles are right-angled, and the two shorter sides are a and b,Sum of two sides = a+b Area = (ab)/2

If the two are equal, then a+b = (ab)/2

Since the numbers involved are only small integers, substitute in different possible values for b. Rearrange to find the matching a. If a falls in the given range, then that makes a correct pair.

For example:

let b=1

a+1 = a/2

2a+2 = a

2a = a-2

a = -2

-2 isn't in the range of 1 to 15, so this isn't a correct pair.

The first pair is at b=3:

a+3 = (3a)/2

2a+6 = 3a

6 =a

Therefore the first pair is a=3, b=6, and the sum/area is 9.

The second pair is at b=4:

a+4 = (4a)/2

2a+8 = 4a

8 = 2a

a = 4

So the second pair is a=4, b=4, and the sum/area is 8.

One other result is when b=6, but of course this is the same as the one above.

Thank you also to Esther who sent in a correct solution although you didn't explain how you worked it out, Esther. Has anybody found another way to solve this problem other than using algebra? Let us know if you have.